1) Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M= mass, L= length , T= time and A= electric current then, A) [ϵ0]=[M−1L−3T2A] B) [ϵ0]=[M−1L−3T4A2] C) [ϵ0]=[M−2L2T−1A−2] D) [ϵ0]=[M−1L2T−1A2] Answer: Option BExplanation:from COULOMB'S law F=14πϵ0q1q2R2 ϵ0=q1q24πFR2 substituting the units we have , ϵ0=C2N.m2=[AT]2[MLT−2][L2]= [M−1L−3T4A2]